You have found the following ages (in years) of all 6 sloths at your local zoo: $ 11,\enspace 10,\enspace 2,\enspace 17,\enspace 13,\enspace 15$ What is the average age of the sloths at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 sloths at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{11 + 10 + 2 + 17 + 13 + 15}{{6}} = {11.3\text{ years old}} $ Find the squared deviations from the mean for each sloth. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $11$ years $-0.3$ years $0.09$ years $^2$ $10$ years $-1.3$ years $1.69$ years $^2$ $2$ years $-9.3$ years $86.49$ years $^2$ $17$ years $5.7$ years $32.49$ years $^2$ $13$ years $1.7$ years $2.89$ years $^2$ $15$ years $3.7$ years $13.69$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.09} + {1.69} + {86.49} + {32.49} + {2.89} + {13.69}} {{6}} $ $ {\sigma^2} = \dfrac{{137.34}}{{6}} = {22.89\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{22.89\text{ years}^2}} = {4.8\text{ years}} $ The average sloth at the zoo is 11.3 years old. There is a standard deviation of 4.8 years.